How will this code will work on a big endian machi

问题:

If I have the code:

uint64_t a = 0x1111222233334444;
uint32_t b = 0;
b = a;
printf("a is %llx ",a);
printf("b is %x ",b);

and the output is :

 a is 1111222233334444 b is 33334444

Questions :

1. Will the behavior be same on big-endian machine?

2. If I assign a's value in b or do a typecast will the result be same in big endian?

回答1:

The code you have there will work the same way. This is because the behavior of downcasting is defined by the C standard.

However, if you did this:

uint64_t a = 0x0123456789abcdefull;
uint32_t b = *(uint32_t*)&a;
printf("b is %x",b)

Then it will be endian-dependent.

EDIT:

Little Endian: b is 89abcdef

Big Endian : b is 01234567

回答2:

When assigning variables, compiler handles things for you, so result will be the same on big-endian.

When typecasting pointers to memory, result will NOT be the same on big-endian.

回答3:

direct assignment will yield the same result on both little endian and big endian.

memory typecast on big endian machine will output

a is 1111222233334444 b is 11112222